Question

Given that \(X=(X_1\ X_2\ X_3)' \sim N_3(\mu,\Sigma)\), where \(\mu=(0\ 0\ 0)'\) and \(\Sigma=\begin{pmatrix}1 & 1 & 0\\1 & 4 & 1\\0 & 1 & 4\end{pmatrix}\). The correlation coefficient between \(X_2\) and \(X_3\) is given by:

• A. 0.25
• B. 0.5
• C. 1
• D. -1

Hint:

Correlation = Cov(X2,X3) divided by the square root of the product of their variances, all read straight off \(\Sigma\).

Answer 1

The Correct Option is A

Step 1: Since $\text{Var}(X_2)=4$ and $\text{Var}(X_3)=4$, the standard deviations are $\sigma_2=\sigma_3=2$. Define standardized variables $Z_2 = X_2/2$ and $Z_3=X_3/2$.
Step 2: Correlation is invariant under positive rescaling, so $\rho_{X_2X_3}=\rho_{Z_2Z_3}=\text{Cov}(Z_2,Z_3)$, since $Z_2,Z_3$ each have unit variance and their covariance directly equals their correlation.
Step 3: Now $\text{Cov}(Z_2,Z_3) = \text{Cov}(X_2/2, X_3/2) = \dfrac{1}{4}\text{Cov}(X_2,X_3) = \dfrac{1}{4}(1) = \dfrac{1}{4}$.
Step 4: Hence $\rho_{X_2X_3} = 0.25$.
\[\boxed{0.25}\]